Determine the force to enable the object to move down

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The discussion centers on calculating the force P required to move a 200kg block down an inclined plane, given a static coefficient of 0.5. The initial calculations led to an incorrect result of P = -1697N, while the expected answer is P = 560N. Key errors identified include miscalculating the components of force P relative to the incline, particularly using the wrong angles for sine and cosine functions. After correcting these mistakes by adjusting the angles, the correct force was determined. The final resolution confirmed the proper calculation method and yielded the expected answer.
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Homework Statement


A force P acts as shown on 200kg block placed on a inclined plane . The static coefficient between the block and the plane is 0.5 . Determine the force P to ensure the block is in the verge of moving down.

Homework Equations

The Attempt at a Solution


Wsin10 +P cos 20 = Fs
P cos 20 = 0.5( P sin20 - W cos 10 ) -Wsin 10
P cos 20 = 0.5( P sin20 - (200x9.81) cos 10 ) -(200x9.81)sin 10
0.17P = 0.5( 0.34P - 1932) - 340.7
0.94P = 0.17P - 966- 340.7
0.17P - 0.94P = 1306.7N
P =-1697N , but the ans given is P = 560N , which part of my working is wrong ?
 

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20° angle is not reference to the inclined plane but to ground level.
 
werson tan said:

Homework Statement


A force P acts as shown on 200kg block placed on a inclined plane . The static coefficient between the block and the plane is 0.5 . Determine the force P to ensure the block is in the verge of moving down.

Homework Equations

The Attempt at a Solution


Wsin10 +P cos 20 = Fs
P cos 20 = 0.5( P sin20 - W cos 10 ) -Wsin 10
P cos 20 = 0.5( P sin20 - (200x9.81) cos 10 ) -(200x9.81)sin 10
0.17P = 0.5( 0.34P - 1932) - 340.7
0.94P = 0.17P - 966- 340.7
0.17P - 0.94P = 1306.7N
P =-1697N , but the ans given is P = 560N , which part of my working is wrong ?
For one thing, the force P is applied 20° above the horizontal. You must calculate the components of P relative to the block on the incline to determine the correct normal force of the block.
 
SteamKing said:
For one thing, the force P is applied 20° above the horizontal. You must calculate the components of P relative to the block on the incline to determine the correct normal force of the block.
P cos 20 = 0.5( P sin20cos10 - (200x9.81) cos 10 ) -(200x9.81)sin 10
0.17P = 0.5( 0.34P - 1932) - 340.7
0.94P = 0.17P - 966- 340.7
0.17P - 0.94P = 1306.7N
P =-1697N , but the ans given is P = 560N , which part of my working is wrong ? my ANS STILL THE SAME
 
werson tan said:
P cos 20 = 0.5( P sin20cos10 - (200x9.81) cos 10 ) -(200x9.81)sin 10
0.17P = 0.5( 0.34P - 1932) - 340.7
0.94P = 0.17P - 966- 340.7
0.17P - 0.94P = 1306.7N
P =-1697N , but the ans given is P = 560N , which part of my working is wrong ? my ANS STILL THE SAME
It should be PCos(20 +10) not PCos(20)...and you to adjust that Sin 20 too.
 
azizlwl said:
It should be PCos(20 +10) not PCos(20)...and you to adjust that Sin 20 too.
THANKS , I NOTED MY MISTAKE , I GT THE ANS FINALLY
 

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